Proof. (Sketch)
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Bounded convex functions automatically have Lipschitz bound on slightly shrinked convex domains.
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The second item follows from a slightly tricky application of mean value inequality for subharmonic functions, cf. [52, section 4.3].
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The third item is because the function has mean value zero, so an upper bound implies an -bound, cf. [52, section 4.3].
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